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The Riemann zeta function $\zeta(s)$ is known to be holomorphic, except at $s=1$.

I am not trained to university maths, but I interpret this to mean:

  • it is complex differentiable at every point (smoothly changing values, no discontinuities)

  • and is equal to Taylor series developed around any point in its domain

Question: How do we show $\zeta(s)$ is holomorphic?

Is it enough to show that the series which represent it, $\zeta(s)=\sum 1/n^s$ valid for $\Re(s)>1$ for example, are infinitely differentiable, except at the known pole at $s=1$?

Is it also correct to say that by the principle of analytic continuation, any other series, such as $\zeta(s)=(1-s^{1-s})^{-1}\eta(s)$ valid over larger domains, are also holomorphic, except at any new poles. By separate analysis, there are no new poles in $0<\Re(s)\leq 1$.

Apologies if this question seems naive, I am self-teaching.

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  • $\begingroup$ Recommendation: Read about analytic continuation: en.wikipedia.org/wiki/Analytic_continuation and then come back to ask about anything still unclear. $\endgroup$
    – GEdgar
    Commented Sep 5, 2021 at 0:45
  • $\begingroup$ hi @GEdgar I know you're trying to help, but that doesn't help me. Is it not as simple as proving the series is infinitely differentiable, except at one pole s=1? $\endgroup$
    – Penelope
    Commented Sep 5, 2021 at 3:12
  • $\begingroup$ Start by recalling local uniform limit of holomorphic is holomorphic, and the usual definition of $\zeta$. Extend that to, e.g., the Dirichlet series $(s-1)\zeta(s)=\sum_{n=1}^\infty\left(\frac{n}{(n+1)^s}-\frac{n-s}{n^s}\right)$ and then use reflection formula. $\endgroup$ Commented Sep 5, 2021 at 5:11
  • $\begingroup$ "Is it not as simple as..." No, it is not. $\endgroup$
    – GEdgar
    Commented Sep 5, 2021 at 12:36
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    $\begingroup$ @Tariq That argument shows that $\zeta(s)$ is holomorphic for $\Re (s)>1$. $\endgroup$
    – Gary
    Commented Sep 5, 2021 at 15:48

1 Answer 1

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Let $U=\{s\in\Bbb{C}\,:\, \text{Re}(s)>1\}$, and consider first $\zeta_0:U\to\Bbb{C}$ defined as \begin{align} \zeta_0(s)&:=\sum_{n=1}^{\infty}\frac{1}{n^s} \end{align} For each $\sigma>1$, if $\text{Re}(s)\geq \sigma$, then \begin{align} \sum_{n=1}^{\infty}\left|\frac{1}{n^s}\right|&=\sum_{n=1}^{\infty}\frac{1}{n^{\text{Re}(s)}}\leq \sum_{n=1}^{\infty}\frac{1}{n^{\sigma}}<\infty \end{align} This implies local uniform convergence of a series of holomorphic functions on $U$, and thus $\zeta_0$ is a holomorphic function on $U$. The only reason I use the notation $\zeta_0$ is to emphasize that we have not done any analytic continuation yet, so it's not the full Riemann zeta function yet. One has the following theorem:

Theorem.

There exists an analytic/holomorphic function $\zeta: \Bbb{C}\setminus \{1\}\to\Bbb{C}$ such that $\zeta|_U=\zeta_0$, and such that $s=1$ is a simple pole of $\zeta$ (with residue $1$).

We call the full mapping $\zeta$ the Riemann zeta function. $\zeta_0$ is merely a part of the restriction. As mentioned in the theorem, one has to prove that such an analytic/holomorphic function $\zeta:\Bbb{C}\setminus\{1\}\to\Bbb{C}$ exists. This is not a trivial thing (in general, existence of analytic continuations/extensions is a non-trivial issue), but in this case there are several well-known proofs of this theorem.

In the first part of my answer, showing the local uniform convergence of the series only shows that $\zeta_0$ is holomorphic on $U$. This says nothing about the existence of an analytic/holomorphic extension of $\zeta_0$ from the initial domain $U$ to $\Bbb{C}\setminus\{1\}$.


As I mentioned, there are several well-known proofs of the theorem. Here are two I remember off the top of my head:

  • The first method can be found in Ahlfors. Start by observing that for $\text{Re}(s)>1$ (i.e $s\in U$) we have $\Gamma(s)\zeta_0(s)=\int_0^{\infty}\frac{x^{s-1}}{e^x-1}\,dx$. One can then deform the integration to get it as an integral over a certain contour $C$ which avoids the non-negative real axis, and then with a bit of work, one can show \begin{align} \zeta_0(s)&=-\frac{\Gamma(1-s)}{2\pi i}\int_C\frac{(-z)^{s-1}}{e^z-1}\,dz \end{align} (see Ahlfors for the details). Now, one observes that on the RHS, the integral is an entire function of $s$ (this is not trivial, but also not too hard if we have things like Fubini's theorem and Morera's theorem available). $\Gamma(1-s)$ has poles at $s=1,2,3,\dots$. In short, the RHS is a mermomorphic function on $\Bbb{C}$; but actually on the LHS, we already know $\zeta_0$ is holomorphic at $s=2,3,4,\dots$. In summary, the RHS provides us with a meromorphic extension of $\zeta_0$ to $\Bbb{C}$ with only a simple pole at $s=1$.

  • Another approach for showing the existence of an analytic continuation involves using Mellin transforms and the Jacobi theta identity. Wikipedia gives a formula for $\zeta_0$. You can find more details in the delightful little book A Brief Introduction to Theta Functions (there are sections on Jacobi's identity, the Mellin Transform and the zeta function).

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  • $\begingroup$ thanks @peek-a-boo. Your answer is clear for $\sigma>1$ but in my self-teaching I keep finding references to the need for the uniform convergence of the series to be on a compact subset of the complex domain. Why is this not an issue here? What am I misunderstanding. Once I've understood this I will mark this as solved. $\endgroup$
    – Penelope
    Commented Jan 8, 2022 at 0:59
  • $\begingroup$ @Tariq here we have a special situation where we have uniform convergence on a closed but not bounded set. This is an extra benefit. i.e we have something better than just uniform convergence on compact sets, so I'm not sure why you think this is a problem. $\endgroup$
    – peek-a-boo
    Commented Jan 8, 2022 at 1:23
  • $\begingroup$ my apologies. We've established uniform convergence on $\sigma>1$ which is better than compact sets. However I can't find the logical step from this to say the series is holomorphic. My interpretation of the textbooks is that the link can only be made if the convergence is on a compact set. This is the source of my puzzle. $\endgroup$
    – Penelope
    Commented Jan 8, 2022 at 15:05

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